sin
Question:

sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9 =

(a) 1

(b) 4

(c) 2

(d) 0

Solution:

(c) 2

We have:

$\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{\pi}{9}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{4 \pi}{9}$

$=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{8 \pi}{18}$

$=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2}\left(\frac{7 \pi}{18}\right)+\sin ^{2}\left(\frac{8 \pi}{18}\right)$

$=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2}\left(\frac{\pi}{2}-\frac{2 \pi}{18}\right)+\sin ^{2}\left(\frac{\pi}{2}-\frac{\pi}{18}\right)$

$=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\cos ^{2} \frac{2 \pi}{18}+\cos ^{2} \frac{\pi}{18}$

$=\sin ^{2} \frac{\pi}{18}+\cos ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\cos ^{2} \frac{2 \pi}{18}$

$=1+1$

= 2

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